Question

A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generated a wave of intensity $l_0$ which is equally divided into two parts : One part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed. The maximum value of $\lambda$ to produce a minima at D is given by :

• A. $\pi R$
• B. $2\pi R$
• C. $\frac{\pi R}{2}$
• D. $\frac{3\pi R}{2}$

Answer 1

The correct option is B $2\pi R$

Net path difference between the waves travelling through path $I$ and $II$, $x=\frac{3}{4}2\pi R-\frac{1}{4}2\pi R$

$⟹x=\pi R$

Let $\lambda$ be the wavelength of the waves.

Now for minima to occur at D, path difference $=(n+\frac{1}{2})\lambda$ where $n=0,1,2,.........$

Thus $\frac{(2n+1)\lambda }{2}=\pi R⟹\lambda =\frac{2\pi R}{2n+1}$

Now wavelength is maximum for $n=0$

Thus ${\lambda }_{max}=2\pi R$